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Sandwich Theorem — Definition, Formula & Examples

The Sandwich Theorem states that if a function is trapped between two other functions that both approach the same limit at a point, then the trapped function must also approach that same limit.

If g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) for all xx in some open interval containing cc (except possibly at cc itself), and lim⁡x→cg(x)=lim⁡x→ch(x)=L\lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, then lim⁡x→cf(x)=L\lim_{x \to c} f(x) = L.

Key Formula

If g(x)≤f(x)≤h(x) and lim⁡x→cg(x)=lim⁡x→ch(x)=L, then lim⁡x→cf(x)=L.\text{If } g(x) \le f(x) \le h(x) \text{ and } \lim_{x \to c} g(x) = \lim_{x \to c} h(x) = L, \text{ then } \lim_{x \to c} f(x) = L.
Where:
  • f(x)f(x) = The function whose limit you want to find
  • g(x)g(x) = The lower bounding function
  • h(x)h(x) = The upper bounding function
  • cc = The point at which the limit is evaluated
  • LL = The common limit of the bounding functions

How It Works

You use the Sandwich Theorem when you cannot evaluate a limit directly but can bound the function above and below by simpler functions whose limits you already know. First, find a lower bound g(x)g(x) and an upper bound h(x)h(x) such that g(x)≤f(x)≤h(x)g(x) \le f(x) \le h(x) near the point of interest. Then compute lim⁡g(x)\lim g(x) and lim⁡h(x)\lim h(x). If both limits equal the same value LL, the theorem guarantees lim⁡f(x)=L\lim f(x) = L. The name comes from the image of ff being "sandwiched" between gg and hh.

Worked Example

Problem: Find lim⁡x→0x2sin⁡ ⁣(1x)\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right).
Establish bounds: Since −1≤sin⁡(1/x)≤1-1 \le \sin(1/x) \le 1 for all x≠0x \neq 0, multiply through by x2x^2 (which is non-negative):
−x2≤x2sin⁡ ⁣(1x)≤x2-x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2
Evaluate the bounding limits: Compute the limits of the lower and upper bounds as x→0x \to 0:
lim⁡x→0(−x2)=0andlim⁡x→0x2=0\lim_{x \to 0} (-x^2) = 0 \quad \text{and} \quad \lim_{x \to 0} x^2 = 0
Apply the Sandwich Theorem: Both bounding functions converge to 0, so the squeezed function shares that limit:
lim⁡x→0x2sin⁡ ⁣(1x)=0\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0
Answer: The limit is 00.

Why It Matters

The Sandwich Theorem is essential in Calculus I for proving foundational results like lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. It also appears in real analysis courses when establishing convergence of sequences and in physics when bounding oscillatory quantities.

Common Mistakes

Mistake: Using bounding functions whose limits are not equal to each other.
Correction: The theorem only applies when both lim⁡g(x)\lim g(x) and lim⁡h(x)\lim h(x) exist and are the same value LL. If the bounds converge to different values, the theorem gives no conclusion.

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